How does Lorentz transformation manage to maintain the speed of light constant?

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Let's say, there are two observers A and B moving with respect to each other with relative velocity v. Now, when we say, the speed of light is same for every observer, does it mean that the speed of light measured by A and B in their own frames, as well as the other's frame will be same?
That means, If A tries to figure out, what will be velocity of light in B's frame, as observed by B, will he come to the answer c (ca. 300,000 km/s), or some different answer?
The reason to ask this is:
Let's say there is a long rod (300,000 km) in B's frame, which B uses to determine speed of light (by checking the time, 1 s, for light to go from one end to the other). Now, A and B being in relative motion, A will see B's rod contracted, while B's time dilated, which will tell him that the speed of the light is decreased in B's frame by some Lorentz multiple? Is that correct?
 

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  • #2
Doc Al
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Let's say, there are two observers A and B moving with respect to each other with relative velocity v. Now, when we say, the speed of light is same for every observer, does it mean that the speed of light measured by A and B in their own frames, as well as the other's frame will be same?
That means, If A tries to figure out, what will be velocity of light in B's frame, as observed by B, will he come to the answer c (ca. 300,000 km/s), or some different answer?
The speed of light being the same for all observers (thus, an invariant) means that any observer will measure the speed of light with respect to himself to be c.
The reason to ask this is:
Let's say there is a long rod (300,000 km) in B's frame, which B uses to determine speed of light (by checking the time, 1 s, for light to go from one end to the other). Now, A and B being in relative motion, A will see B's rod contracted, while B's time dilated, which will tell him that the speed of the light is decreased in B's frame by some Lorentz multiple? Is that correct?
No, not correct. You forgot about the relativity of simultaneity. Frame A will disagree that B's clocks will have changed by 1 second (of B time) while the light traveled from one end to the other of the rod. When you account for all three "effects" (which are contained in the Lorentz transformations)--time dilation, length contraction, and clock desynchronization--you will see that A will measure the speed of the light to be c, as always.
 
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The speed of light being the same for all observers (thus, an invariant) means that any observer will measure the speed of light with respect to himself to be c.
If one finds the speed of light to be c w.r.t himself, that is his own frame, but not in other's frame, then it is sufficient. No question arise.

No, not correct. You forgot about the relativity of simultaneity. Frame A will disagree that B's clocks will have changed by 1 second (of B time) while the light traveled from one end to the other of the rod. When you account for all three "effects" (which are contained in the Lorentz transformations)--time dilation, length contraction, and clock desynchronization--you will see that A will measure the speed of the light to be c, as always.
Sure, but it will add to dilated time only, and thus observed velocity will decrease once more, No? By the way, I didn't forgot, but It didn't came to me at all!:cry:

Edit: To remove relativity of simultaneity, we may reduce the length of the rod to 150,000 km, and account for two way journey, so we will be worried about only one clock and one rod in B's frame.
 
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  • #4
Doc Al
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Sure, but it will add to dilated time only, and thus observed velocity will decrease once more, No?
No, just the opposite.
By the way, I didn't forgot, but It didn't came to me at all!:cry:
The relativity of simultaneity is crucial to understanding relativity.

You really need to get a book!
 
  • #5
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The reason to ask this is:
Let's say there is a long rod (300,000 km) in B's frame, which B uses to determine speed of light (by checking the time, 1 s, for light to go from one end to the other). Now, A and B being in relative motion, A will see B's rod contracted, while B's time dilated, which will tell him that the speed of the light is decreased in B's frame by some Lorentz multiple? Is that correct?
Let us say, A measures that B is travelling at 250000 km/s. And let us call the ends of the rod R1 and R2.
In B's frame light starts at the space-time coordinates R1'=0, t1'=0 and arrives at R2'=300000km, t2'=1s.

To find out the coordinates in A's frame, we use the Lorentz transform. This brings the result in A's frame:
Start: R1=0, t1=0; and arrival: R2=994987,437..km, t2=3,317..s.
Because velocity is defined as the rate of change of position, we calculate R2/t2=994987,437../3,317..=300000 km/s.
This means also in A the speed of light has the same value.

PS: You should listen to Doc Al: It's all about relativity of simultaneity.:smile:
 
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  • #6
robphy
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One answer to the title of this thread
"how does Lorentz transformation manage to maintain the speed of light constant?"
is that
two eigenvectors of the Lorentz [boost] Transformations are lightlike 4-vectors.
 
  • #7
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One answer to the title of this thread
"how does Lorentz transformation manage to maintain the speed of light constant?"
is that
two eigenvectors of the Lorentz [boost] Transformations are lightlike 4-vectors.
Exactly!
The best possible answer one can give.
 

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