Personality Cafe banner

1 - 20 of 30 Posts

·
Premium Member
I thimk INTP
Joined
·
14,613 Posts
Discussion Starter #1
Huh? Does anyone understand what this guy is saying? Any clue would be a start. He talks faster than I can take it in yet he seems so confident he must be right ...

What if photos do have a little mass ... just because they are light? What does the Lorentz transformation have to do with anything? What does "speed of causality mean? What other things are we assuming here? Is the present value of c just a fluke?

 

·
Registered
Joined
·
1,739 Posts
What if photos do have a little mass ... just because they are light?
If photons had mass, they could not move at the speed of light. Any massive object moving at that speed would have infinite energy. They do however have momentum despite being massless.

What does the Lorentz transformation have to do with anything?
The Lorentz transformation accounts for relativistic effects on velocity in translating between frames of reference. If you see a vehicle moving at 0.6c shoot a bullet at 0.6c relative to itself, the speed of the bullet relative to you won't be 1.2c, but rather some sub-c value. Lorentz transformations allow you to calculate that value.

What does "speed of causality mean?
Speed of causality refers to the spacetime cone. If two events cannot be linked by a ray of light, then one cannot cause the other, and what's more there exists a frame of reference in which those events are simultaneous; we call this relation space-like. Alternatively you can have a time-like relation, wherein one event precedes another in every frame of reference. It is in a time-like relation that causality is possible.

What other things are we assuming here?
Assuming lots of things, for sure. A big one is that the speed of light in a vacuum is constant, which is itself a consequence of the assumption of the isometry of space. One might think to note the frame-invariance of the speed of light (that is there exists no frame in which the speed of light equals something other than c), though that has been experimentally verified by Michelson. Something I should mention here is that special relativity (i.e. the topic that this video touches upon) assumes that all frames of reference are inertial; if you want to deal with non-inertial frames, you need general relativity.

Is the present value of c just a fluke?
The exact value of the speed of light is most likely arbitrary. That there exists such a speed is a far more interesting observation than what that speed is.
 

·
Premium Member
I thimk INTP
Joined
·
14,613 Posts
Discussion Starter #3
Your explanations seem helpful though I cannot verify them. I'll just ask about the first part.
If photons had mass, they could not move at the speed of light. Any massive object moving at that speed would have infinite energy. They do however have momentum despite being massless.
I thought momentum = mass x velocity. If mass = 0, are we still talking about classical Newtonian physics? Do photons have an impact when they strike something? Why not photos with a very tiny mass? Then tiny mass x large energy (not infinite) = c.

Wait a minute. If E = mc[SUP]2[/SUP], then if photos have energy, and m = E/c2. This energy is far from infinite.
 

·
Registered
Joined
·
1,739 Posts
Your explanations seem helpful though I cannot verify them. I'll just ask about the first part.
I thought momentum = mass x velocity. If mass = 0, are we still talking about classical Newtonian physics? Do photons have an impact when they strike something? Why not photos with a very tiny mass? Then tiny mass x large energy (not infinite) = c.

Wait a minute. If E = mc[SUP]2[/SUP], then if photos have energy, and m = E/c2. This energy is far from infinite.
p = mv is a classical approximation of momentum. We are not talking about classical Newtonian physics. If photons had mass, then that would imply that an object with mass could reach the speed of light. Experiments in particle physics strongly indicate that this is not the case.

Likewise, E = mc[sup]2[/sup] is an approximation in the low-momentum limit (sometimes it will be called resting energy). The general form of the equation is E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]. For photons, m = 0 and therefore E = pc.
 

·
Registered
Joined
·
2,208 Posts
I don't buy into non-mechanical interpretations of photon motion which say things like that photons have no mass or no spatial extension (point-particles).

Mass is a function of density and volume and photons have both density and volume. Photons have spatial extension (i.e. a finite radius) and they exert radiation pressure. You can set microscopic objects into rotation by directed photon bombardment.
 

·
Banned
Joined
·
3,344 Posts
I don't buy into non-mechanical interpretations of photon motion which say things like that photons have no mass or no spatial extension (point-particles).

Mass is a function of density and volume and photons have both density and volume. Photons have spatial extension (i.e. a finite radius) and they exert radiation pressure. You can set microscopic objects into rotation by directed photon bombardment.
Photons have no rest mass because they are never at rest it any inertial frame. Photons however do have relativistic mass which is the same in every inertial frame E = hf = mc^2 therefore m = hfc^-2
 

·
Banned
Joined
·
3,344 Posts
Could you define the photon as being at rest and say everything else is moving relative to the stationary photon?
Then it wouldn't be a photon anymore. From the perspective of a photon nothing moves, since there is no time nor space, for it it is all contracted into a point. Even saying it has a perspective is assuming it has a rest frame which it doesn't, so it doesn't actually even make sense.
 
  • Like
Reactions: PiT

·
Registered
Joined
·
2,208 Posts
From the perspective of a photon nothing moves,
How can it simultaneously be the case that a photon travels relative to me but I don't travel relative to the photon?

since there is no time nor space, for it it is all contracted into a point. Even saying it has a perspective is assuming it has a rest frame which it doesn't, so it doesn't actually even make sense..
If a photon is moving then it experiences time. If there is motion happening then time is happening.

If I define the photon itself as the frame of reference then it experiences itself as stationary but there's still a background which moves relative to the photon.
 

·
Banned
Joined
·
3,344 Posts
How can it simultaneously be the case that a photon travels relative to me but I don't travel relative to the photon?
Answer below:


If a photon is moving then it experiences time. If there is motion happening then time is happening.

If I define the photon itself as the frame of reference then it experiences itself as stationary but there's still a background which moves relative to the photon.
A photon is not moving relative to itself. Motion is relative, as is time. The time between two events depends on the reference frame within which it is measured, and for a photon it is always 0. Time does not exist for a photon, but all events are simultaneous. Since they are all simultaneous, there is no motion.
 

·
Registered
Joined
·
2,208 Posts
Motion is relative, as is time.
Right, so if the photon is moving relative to me then it can also be said that I'm moving relative to the photon. From the photon's point of view I'm moving but it is not.

I can either make the photon the stationary reference frame or I can make myself the stationary reference frame.

The time between two events depends on the reference frame within which it is measured, and for a photon it is always 0. Time does not exist for a photon, but all events are simultaneous. Since they are all simultaneous, there is no motion.
Earlier this year at the LHC they detected photons colliding and bouncing off of other photons. From any reference frame outside the photon it appears the photon is experiencing collisions and changing direction. From the photon's point of view the orientation of the background changed with each collision.

I don't see how it logically follows that motion would cease to exist for a photon. I'm not moving relative to myself but motion outside of me still exists relative to my stationariness. I can't make any coherent sense of the notion that a body moving at a finite velocity could experience all events simultaneously.
 

·
Premium Member
I thimk INTP
Joined
·
14,613 Posts
Discussion Starter #12
I don't see how it logically follows that motion would cease to exist for a photon. I'm not moving relative to myself but motion outside of me still exists relative to my stationariness. I can't make any coherent sense of the notion that a body moving at a finite velocity could experience all events simultaneously.
What about this? A photon is a special particle/wave. It moves at the maximum possible. From our POV everything else moves very slowly. From the photon's POV everything moves so slowly it can think of other things as not moving at all. Also if a photon doesn't experience time or rather zero time passes, everything else wouldn't move. Even for us, if zero time passes, other things don't get the chance to move. Second thought. If we had the math for the photon in space/time maybe this could be more clear. As it is we are using only words and words don't capture the details. Better?
 

·
Registered
Joined
·
2,208 Posts
From the photon's POV everything moves so slowly it can think of other things as not moving at all. Also if a photon doesn't experience time or rather zero time passes, everything else wouldn't move. Even for us, if zero time passes, other things don't get the chance to move. Second thought. If we had the math for the photon in space/time maybe this could be more clear. As it is we are using only words and words don't capture the details. Better?
Wouldn't it be the case that, motion being relative, things are moving relative to the photon at exactly the same speed as the photon is moving relative to those things?

I am moving relative to the photon exactly as fast as the photon is moving relative to me.

A photon traveling away from me at c is the same as saying the photon is standing still and I am moving away from the stationary photon at c.
 

·
Premium Member
I thimk INTP
Joined
·
14,613 Posts
Discussion Starter #14
A photon is not moving relative to itself. Motion is relative, as is time.
That is true for us also. We are the center of the universe and don't move relative to ourselves. But can two photons smash into each other? What does that mean? Does one of them disappear? =====================================




Wouldn't it be the case that, motion being relative, things are moving relative to the photon at exactly the same speed as the photon is moving relative to those things? I am moving relative to the photon exactly as fast as the photon is moving relative to me. A photon traveling away from me at c is the same as saying the photon is standing still and I am moving away from the stationary photon at c.
I am guessing that we can't speak of moving relative to the photon at all because the photon is moving so fast we are relatively stationary. It must have to do with the curvature of space/time and I don't have the equations.
 

·
Registered
Joined
·
1,739 Posts
Wouldn't it be the case that, motion being relative, things are moving relative to the photon at exactly the same speed as the photon is moving relative to those things?

I am moving relative to the photon exactly as fast as the photon is moving relative to me.

A photon traveling away from me at c is the same as saying the photon is standing still and I am moving away from the stationary photon at c.
You're trying to think about this in classical terms. In special relativity, there must exist a Lorentz transformation between two frames of reference for you to be able to relate them accounting for relativistic effects. There exists no Lorentz transformation between the hypothetical photonic rest frame and your own rest frame.

For this reason, you cannot simply relate what you see to what the photon would see. To do so would constitute an undefined operation, akin to dividing by zero.
 

·
Registered
Joined
·
1,739 Posts
That is true for us also. We are the center of the universe and don't move relative to ourselves. But can two photons smash into each other? What does that mean? Does one of them disappear?
I just noticed this, but photons are bosonic matter, as opposed to almost everything else we know in our daily lives which is fermionic. Bosons can freely occupy the same space, so there is no issue here. Two photons would pass through each other. They can interfere because of their wave properties, but they won't smash like the particles you see in the LHC.

Can you elaborate on what you mean? Are you able to provide a mechanical argument?
I am unsure of how best to approach such a project since this is a quite theoretical topic, but I can try. @Euclid referred to the concept of relativistic mass (which is a different way of formulating the situation than I am accustomed to, but maybe it will help to use it here). For an object with mass, relativistic mass grows proportionally to rest mass as the object accelerates. If it helps to conceptualize it, relativistic mass = rest mass * (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]-1/2[/sup].

As the object approaches c, relativistic effects cause its effective mass to diverge (that is, approach an infinite number). The amount of energy needed for an object with rest mass to reach c is infinite. The Lorentz transformation also uses this (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]-1/2[/sup] factor (known as gamma), so it runs into the same problems.

As you approach c, length contracts and time dilates as a consequence of this as well as the non-simultaneity of events under transformations. To expand on this concept, if you were to measure the length of an object moving past you at relativistic speeds, you would measure its length as being shorter than what it would be in that object's rest frame. Your ruler might touch both ends at the same time in your own frame, but it would not be simultaneous in that rest frame, and that introduces a measurement error.

Following from the equations of the Lorentz transformation, the photons exist in a frame relative to us where gamma is undefined. Their world has no length and time is dilated infinitely. Boosting into this frame is a theoretical absurdity, and we cannot speak to it with any sense of rigor.
 

·
Banned
Joined
·
3,344 Posts
Can you elaborate on what you mean? Are you able to provide a mechanical argument?
The Lorentz transformation defines how measurements in different inertial reference frames relate to each other. If you walk inside a train at a velocity A, while the train itself travels at a velocity B, in galilean tranformation The speed you are walking relation to the ground would be A+B, but according to Lorentz transformation this is not true but is merely an approximation. The faster you travel, and the faster the train travels, the less accurate it becomes. Furthermore events that occur simultaneous outside the train are no longer simultaneous viewed from inside the train and vice versa. The head of the train will trail in the past while the tail end tilts into the future. The mass of the train also increases as it increases velocity and the movement inside the train slows down. Now from the perspective of the train it is the other way around, the events outside the train are slowing down, so how can these two things be reconciled? The answer to this is that the events outside the train are also timeshifted, just as the events inside the train are timeshifted from the perspective from outside the train.


As you approach the speed of light mass also approaches infinity, all length contracts toward a single point and time freezes.
 

·
Registered
Joined
·
2,208 Posts
If it helps to conceptualize it, relativistic mass = rest mass * (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]-1/2[/sup].
The Lorentz transformation also uses this (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]-1/2[/sup] factor (known as gamma), so it runs into the same problems.
And from whose frame of reference is the Lorentz transformation derived? Where am I looking from?
 

·
Banned
Joined
·
3,344 Posts
And from whose frame of reference is the Lorentz transformation derived? Where am I looking from?
The Lorentz transformation holds for any two reference frames where v < c, but is undefined if v = c. If you try to put in v = c then you end up trying to divide by zero which you can't.
 
1 - 20 of 30 Posts
Top