Will the velocity of light be the same in all reference frames?

[Aswin Sasikumar 1729]

If yes then imagine what I am going to say...
From a source two photons are emmutted symultaniously.
If one of the photon had eyes to see what will 'he ' measures the velocity of the other photon which is moving with 'him'? Won't it be zero?!

 

[Demystifier]

By "all" reference frames, one usually means only those which move slower than $c$. This condition is not satisfied in your case.

 

[Ibix]

There isn't a reference frame in which light is at rest. The notion is contradictory, as you yourself noted, since light would have to be both at rest and moving c.

So the answer is that it isn't possible to describe what a light pulse would see. The question doesn't even makes sense in relativity.

 

[Dale]

Note that the correct statement is that the speed of light is the same in all inertial reference frames. It can be different in non inertial frames.

 

[robphy]

Note that the correct statement is that the speed of light is the same in all inertial reference frames. It can be different in non inertial frames.

One would have to specify the definition of "speed".

From a given event, light travels on the lightcone, regardless of the reference frame (inertial or non inertial).

[I should clarify that this is a statement about the tangent space at that event... a space of direction vectors in space-time from that event.]

 

[Aswin Sasikumar 1729]

Note that the correct statement is that the speed of light is the same in all inertial reference frames. It can be different in non inertial frames.

Sir will u please explain about the non-inertual frames?

 

[Aswin Sasikumar 1729]

By "all" reference frames, one usually means only those which move slower than $c$. This condition is not satisfied in your case.

A law must be a law.So 'ALL' should mean all itself.

 

[DrGreg]

A law must be a law.So 'ALL' should mean all itself.

The correct statement should be that the speed of light is the same in all inertial reference frames. A frame in which light is at rest is not inertial.

 

[Dale]

Sir will u please explain about the non-inertual frames?

A typical example is a rotating reference frame where the speed of light can be substantially different from c at great distances from the axis. The restriction on c is for inertial frames.

 

[Demystifier]

A law must be a law.So 'ALL' should mean all itself.

For all numbers $x$ there is a number $y$ such that $xy=1$. This is a mathematical law. Yet, there is an exception: $x=0$.

The reason why velocity of light is an exception has the same mathematical origin as the purely mathematical example above. In both cases there is a division by zero involved.

 

[Demystifier]

Some people above suggested that there is no reference frame in which light is at rest, or that such a frame is not inertial. Both statements are wrong. There is such a frame and it is an inertial frame. But such a frame is not a Lorentz frame. It is a light-cone frame:
https://en.wikipedia.org/wiki/Light-cone_coordinates

 

[Dale]

Some people above suggested that there is no reference frame in which light is at rest, or that such a frame is not inertial. Both statements are wrong. There is such a frame and it is an inertial frame. But such a frame is not a Lorentz frame. It is a light-cone frame:
https://en.wikipedia.org/wiki/Light-cone_coordinates

Light cone coordinates are not an inertial frame. By the second postulate light travels at c in all inertial frames and light does not travel at c in light cone coordinates. Furthermore, the coordinate basis of light cone coordinates do not even define a valid reference frame since a reference frame, by definition, consists of three spacelike and one timelike orthonormal vector fields.

 

[Demystifier]

Light cone coordinates are not an inertial frame. By the second postulate light travels at c in all inertial frames and light does not travel at c in light cone coordinates. Furthermore, the coordinate basis of light cone coordinates do not even define a valid reference frame since a reference frame, by definition, consists of three spacelike and one timelike orthonormal vector fields.

Hm, it seems that you and me are using different definitions of the word "frame". In my "dictionary", the frame is the same as the system of coordinates. In your "dictionary", frames are a subclass of systems of coordinates.

Apart from this terminology difference, I think there is no any reason to think of light-cone coordinates as non-inertial coordinates. They are inertial, in the sense that lines of the coordinate mesh are geodesics. Indeed, light moves along one of such lines, and such a motion of light is inertial in the sense that no force (except gravity) influences it.

 

[Dale]

Hm, it seems that you and me are using different definitions of the word "frame". In my "dictionary", the frame is the same as the system of coordinates. In your "dictionary", frames are a subclass of systems of coordinates.

In standard GR terminology a reference frame is called a frame field, tetrad, or vierbein https://en.m.wikipedia.org/wiki/Frame_fields_in_general_relativity

It is a set of four vector fields, three of which are spacelike, one of which is timelike, and all of which are orthonormal to each other.

A coordinate system is not a set of four vector fields, but the distinction can be blurry because you can always use the coordinate basis to generate some vector fields known as the coordinate basis. However, a coordinate basis may not be orthonormal etc. so not all coordinate bases qualify as valid reference frames.

Apart from this terminology difference, I think there is no any reason to think of light-cone coordinates as non-inertial coordinates. They are inertial, in the sense that lines of the coordinate mesh are geodesics. Indeed, light moves along one of such lines, and such a motion of light is inertial in the sense that no force (except gravity) influences it.

While I agree that the coordinate lines are geodesics, that is simply not how a reference frame is defined in relativity.

 

[vanhees71]

A frame is something in the real world. You can take, e.g., a corner of your lab and the three edges of the room as the spatial reference frame and a clock to define a reference frame.

 

[Demystifier]

OK, you convinced me, I will update my dictionary accordingly.

 

[Aswin Sasikumar 1729]

For all numbers $x$ there is a number $y$ such that $xy=1$. This is a mathematical law. Yet, there is an exception: $x=0$.

The reason why velocity of light is an exception has the same mathematical origin as the purely mathematical example above. In both cases there is a division by zero involved.

Sir I am not having knowledge in what u people were discussing.But this above statement u gave is wrong.In fact incomplete .The correct law is :"for all non zero x there exist a y such that xy=1".

 

[Demystifier]

Sir I am not having knowledge in what u people were discussing.But this above statement u gave is wrong.In fact incomplete .The correct law is :"for all non zero x there exist a y such that xy=1".

Right. Likewise, the correct relativity law is: For all inertial frames with non zero $\gamma^{-1}$ the speed of light is the same, where
$$\gamma^{-1}=\sqrt{1-v^2/c^2}$$
and $v$ is the speed of the frame.

 

[Aswin Sasikumar 1729]

Right. Likewise, the correct relativity law is: For all inertial frames with non zero $\gamma^{-1}$ the speed of light is the same, where
$$\gamma^{-1}=\sqrt{1-v^2/c^2}$$
and $v$ is the speed of the frame.

I am convinced with this answer but there is no such word "non zero" in the Einstein's statement which is in our book.Thank u.

 

[Mister T]

If yes then imagine what I am going to say...
From a source two photons are emmutted symultaniously.
If one of the photon had eyes to see what will 'he ' measures the velocity of the other photon which is moving with 'him'? Won't it be zero?!

Let's start with the postulate that the speed of light is independent of the speed of the source. From there you can deduce that such a speed must be the fastest speed possible, and that no observer can travel that fast. Thus your "photon" traveling at the speed of light cannot be replaced with an observer, as observers cannot move that fast. If they could, the premise on which you based your conclusion would have to be false.

 

[robphy]

I am convinced with this answer but there is no such word "non zero" in the Einstein's statement which is in our book.Thank u.

A word of advice...
although Einstein made a remarkable achievement in formulating relativity and revealing many of its non-intuitive implications for physics, his words are not sacred.

It took the work of many folks and many years to try to clarify the meaning of ideas in relativity. For instance, Einstein didn't initially appreciate Minkowski and his mathematical formulation of relativity using "space-time". Later, he came to appreciate it and it helped him formulate General Relativity. Even then, Einstein and others made mistakes in relativity (for example, with the coordinate singularity in the Schwarzschild solution). So, the point is: after Einstein's brilliant papers on the subject, his ideas and definitions have been refined to clarify what is really going on... in order to prevent misunderstandings and misuses... not only in speech and interpretation, but in mathematical theorems.

 

[Dale]

the correct relativity law is: For all inertial frames with non zero $\gamma^{-1}$ ...

there is no such word "non zero" in the Einstein's statement.

The "non-zero" qualifier is not necessary. As I said back in post 4 the correct statement is that the speed of light is the same in all inertial reference frames. In relativity, the term "inertial reference frame" is purposely defined (see post 14) so as to make that statement true.

 

[Aswin Sasikumar 1729]

The "non-zero" qualifier is not necessary. As I said back in post 4 the correct statement is that the speed of light is the same in all inertial reference frames. In relativity, the term "inertial reference frame" is purposely defined (see post 14) so as to make that statement true.

Thank u sooo much

 

[Aswin Sasikumar 1729]

Let's start with the postulate that the speed of light is independent of the speed of the source. From there you can deduce that such a speed must be the fastest speed possible, and that no observer can travel that fast. Thus your "photon" traveling at the speed of light cannot be replaced with an observer, as observers cannot move that fast. If they could, the premise on which you based your conclusion would have to be false.

If the photon has eye, then?

 

[Ibix]

If the photon has eye, then?

What's the eye made out of? The only way we know how to make an eye is out of regular matter that can't move at lightspeed.

It isn't enough to ask "what if...?" You also have to think about whether your hypothetical scenario is consistent with the laws of physics. A photon with eyes isn't.

 

[Aswin Sasikumar 1729]

What's the eye made out of? The only way we know how to make an eye is out of regular matter that can't move at lightspeed.

It isn't enough to ask "what if...?" You also have to think about whether your hypothetical scenario is consistent with the laws of physics. A photon with eyes isn't.

Then I will clear my side more scientifically.With out having relative velocity equal to zero, how can two particle able to move together?

 

[Nugatory]

Then I will clear my side more scientifically.With out having relative velocity equal to zero, how can two particle able to move together?

As asked, that question makes no sense. If two things are moving towards one another and there exists a coordinate system in which one of them is at rest, then they necessarily have a non-zero relative velocity using that coordinate system.

(If there is no such coordinate system, as is the case with flashes of light, then their relative velocity is undefined, although there may exist coordinate systems in which both are moving. In this case they will have a non-zero closing velocity).

 

[Ibix]

Then I will clear my side more scientifically.With out having relative velocity equal to zero, how can two particle able to move together?

Relative velocity can mean two different things. Only one of them has meaning for light.

For an observer such as ourselves we can see two things moving in the same direction with constant separation. We say that their separation rate is zero. In this sense it is meaningful to talk of relative velocity being zero between two light pulses.

In Galilean relativity we could transform to a frame where one of the objects was at rest and find that the other also has velocity zero. You can do the same in Einsteinian relativity - except there are no frames in which light is at rest, so you can't do it for light pulses. In this sense, it is meaningless to talk of a velocity relative to light.

Your intuition that the first relative velocity and the second are the same thing is wrong. You'll never notice the difference at every day speeds, but it is at the root of your problems in this thread.

 

[vanhees71]

The relative velocity for two particles, where at least one particle (say particle 1) has non-zero mass is the velocity of the other particle in the rest frame of particle 1 (which is called the lab frame in scattering theory for historical reasons, when the accelerator experiments where fixed-target experiments). It's easy to formulate this in a manifestly covariant way. In the lab frame we have for the four-momenta (I use natural units, $c=1$)
$$p_1^*=(m_1,0)^T, \quad p_2^*=(E_2^*,\vec{p}_2^*) \; \Rightarrow \; v_{\text{rel}}=\frac{|\vec{p}_2^*|}{E_2^*}=\frac{\sqrt{E_2^{*2}-m_2^2}}{E_2^*}.$$
Now we have for the Minkowski product in this frame and thus for any other frame
$$p_1^* \cdot p_2^*=p_1 \cdot p_2=m_1 E_2^* \; \Rightarrow \; E_2^*=\frac{p_1 \cdot p_2}{m_1},$$
and thus
$$E_2^{*2}-m_2^2=\frac{(p_1 \cdot p_2)^2}{m_1^2}-m_2^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{m_1^2},$$
which leads to
$$v_{\text{rel}}=\frac{\sqrt{(p_1 \cdot p_2)^2-m_1^2 m_2^2}}{p_1 \cdot p_2}.$$
Now, if $m_1=m_2=0$ you have two possibilities: $p_1 \cdot p_2=0$; then the relative velocity is undefined because then $p_1 \propto p_2$ or $p_1 \cdot p_2 \neq 0$, and the relative velocity is 1. If $m_1 \neq 0$ but $m_2=0$ you have always $v_{\text{rel}}=1$.

Also note that $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$. The general expression in an arbitrary frame is
$$v_{\text{rel}}=\frac{1}{1-\vec{v}_1 \cdot \vec{v}_2} \sqrt{(\vec{v}_1-\vec{v}_2)^2-(\vec{v}_1 \times \vec{v}_2)^2}.$$
For a derivation, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.5 and 1.6.

 

[robphy]

Neat.
Should

$$v_{\text{rel}}^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{p_1 \cdot p_2}.$$

be squared in the denominator?

This looks like $\tanh^2\theta=\frac{\sinh^2\theta}{\cosh^2\theta}=\frac{\cosh^2\theta-1}{\cosh^2\theta}$ for nonzero $m$s.

 

[Mister T]

If the photon has eye, then?

Then it can't travel at speed c. Which was your premise. So your argument is self-contradictory.

 

[PAllen]

The relative velocity for two particles, where at least one particle (say particle 1) has non-zero mass is the velocity of the other particle in the rest frame of particle 1 (which is called the lab frame in scattering theory for historical reasons, when the accelerator experiments where fixed-target experiments). It's easy to formulate this in a manifestly covariant way. In the lab frame we have for the four-momenta (I use natural units, $c=1$)
$$p_1^*=(m_1,0)^T, \quad p_2^*=(E_2^*,\vec{p}_2^*) \; \Rightarrow \; v_{\text{rel}}=\frac{|\vec{p}_2^*|}{E_2^*}=\sqrt{E_2^{*2}-m_2^2}{E_2^*}.$$
Now we have for the Minkowski product in this frame and thus for any other frame
$$p_1^* \cdot p_2^*=p_1 \cdot p_2=m_1 E_2^* \; \Rightarrow \; E_2^*=\frac{p_1 \cdot p_2}{m_1},$$
and thus
$$E_2^{*2}-m_2^2=\frac{(p_1 \cdot p_2)^2}{m_1^2}-m_2^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{m_1^2},$$
which leads to
$$v_{\text{rel}}=\frac{\sqrt{(p_1 \cdot p_2)^2-m_1^2 m_2^2}}{p_1 \cdot p_2}.$$
Now, if $m_1=m_2=0$ you have two possibilities: $p_1 \cdot p_2=0$; then the relative velocity is undefined because then $p_1 \propto p_2$ or $p_1 \cdot p_2 \neq 0$, and the relative velocity is 1. If $m_1 \neq 0$ but $m_2=0$ you have always $v_{\text{rel}}=1$.

Also note that $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$. The general expression in an arbitrary frame is
$$v_{\text{rel}}=\frac{1}{1-\vec{v}_1 \cdot \vec{v}_2} \sqrt{(\vec{v}_1-\vec{v}_2)^2-(\vec{v}_1 \times \vec{v}_2)^2}.$$
For a derivation, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.5 and 1.6.

Nice. This is eqivalent to the first formula in a post of mine a while ago: https://www.physicsforums.com/threads/generalized-velocity-addition-and-aberration-formulas.821733/ , where I note that the cross product term is a relativistic correction for orthogonal component of the relative motion, while the denominator not being 1 is a correction for the parallel component.

Also note that, as you are well aware, your statement " $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$" neglects the denominator, which 'implements' parallel velocity addition. Thus, norm of vector difference isn't even true for the parallel case.

 

[Aswin Sasikumar 1729]

As asked, that question makes no sense. If two things are moving towards one another and there exists a coordinate system in which one of them is at rest, then they necessarily have a non-zero relative velocity using that coordinate system.

(If there is no such coordinate system, as is the case with flashes of light, then their relative velocity is undefined, although t

a

The relative velocity for two particles, where at least one particle (say particle 1) has non-zero mass is the velocity of the other particle in the rest frame of particle 1 (which is called the lab frame in scattering theory for historical reasons, when the accelerator experiments where fixed-target experiments). It's easy to formulate this in a manifestly covariant way. In the lab frame we have for the four-momenta (I use natural units, $c=1$)
$$p_1^*=(m_1,0)^T, \quad p_2^*=(E_2^*,\vec{p}_2^*) \; \Rightarrow \; v_{\text{rel}}=\frac{|\vec{p}_2^*|}{E_2^*}=\sqrt{E_2^{*2}-m_2^2}{E_2^*}.$$
Now we have for the Minkowski product in this frame and thus for any other frame
$$p_1^* \cdot p_2^*=p_1 \cdot p_2=m_1 E_2^* \; \Rightarrow \; E_2^*=\frac{p_1 \cdot p_2}{m_1},$$
and thus
$$E_2^{*2}-m_2^2=\frac{(p_1 \cdot p_2)^2}{m_1^2}-m_2^2=\frac{(p_1 \cdot p_2)^2-m_1^2 m_2^2}{m_1^2},$$
which leads to
$$v_{\text{rel}}=\frac{\sqrt{(p_1 \cdot p_2)^2-m_1^2 m_2^2}}{p_1 \cdot p_2}.$$
Now, if $m_1=m_2=0$ you have two possibilities: $p_1 \cdot p_2=0$; then the relative velocity is undefined because then $p_1 \propto p_2$ or $p_1 \cdot p_2 \neq 0$, and the relative velocity is 1. If $m_1 \neq 0$ but $m_2=0$ you have always $v_{\text{rel}}=1$.

Also note that $v_{\text{rel}}$ is only $|\vec{v}_1-\vec{v}_2|$ if $\vec{v}_1 \propto \vec{v}_2$. The general expression in an arbitrary frame is
$$v_{\text{rel}}=\frac{1}{1-\vec{v}_1 \cdot \vec{v}_2} \sqrt{(\vec{v}_1-\vec{v}_2)^2-(\vec{v}_1 \times \vec{v}_2)^2}.$$
For a derivation, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.5 and 1.6.

whether these equations are dimensionally correct?

 

[Ibix]

a

whether these equations are dimensionally correct?

Yes. As @vanhees71 noted he is using natural units in which c=1, and therefore suppressing the factors of c, or including them in the units if you prefer to see it that way.

I do think there's a typo in the last part of the first equation - I think he should be dividing by $E^*_2$ rather than multiplying.

 

[vanhees71]

I do think there's a typo in the last part of the first equation - I think he should be dividing by $E^*_2$ rather than multiplying.

Thanks, I've corrected the typo.