Does c-u mean that c composes with u?

[bernhard.rothenstein]

Consider a source of light S stationary in I and an observer R moviong with speed u . At the origin of time R is located in front of S receiving a first light signal emitted by S. S emits a second light signal at a time Te which is received by R at a time Tr. From the obvious equation in I
c(Tr-Te)=uTr (1)
we obtain
Tr=cTe/(c-u). (2)
Does (2) suggest that c and u compose in the classical way?
sine ira et studio

 

[HallsofIvy]

Consider a source of light S stationary in I and an observer R moviong with speed u . At the origin of time R is located in front of S receiving a first light signal emitted by S. S emits a second light signal at a time Te which is received by R at a time Tr. From the obvious equation in I
c(Tr-Te)=uTr (1)

I assume you mean that R coincides with S initially and is moving AWAY from S at speed u. From t= 0 to Tr, S moves a distance uTr so the light beam must travel that distance: c(Tr- Te)= uTr.

we obtain
Tr=cTe/(c-u). (2)
Does (2) suggest that c and u compose in the classical way?
sine ira et studio

Since everything is done in S's frame of reference, yes. Of course, in R's frame of reference the calculation will be quite different.

 

[bernhard.rothenstein]

c-u

I assume you mean that R coincides with S initially and is moving AWAY from S at speed u. From t= 0 to Tr, S moves a distance uTr so the light beam must travel that distance: c(Tr- Te)= uTr.


Since everything is done in S's frame of reference, yes. Of course, in R's frame of reference the calculation will be quite different.

Thanks. Could you tell me what is the physical meaning of c-u?

 

[JesseM]

Thanks. Could you tell me what is the physical meaning of c-u?

It's sometimes called the "closing velocity"--the speed that the distance between two objects is seen to shrink (or grow, if the faster object is moving away from the slower one) as seen in the frame of a third observer. Another example is that if I see two ships moving towards each other, both moving at 0.6c in my frame, then for me the closing velocity is 1.2c (i.e. if they start out 1.2 light years apart as measured by my ruler, it will take them 1 year to meet according to my clocks), even though each ship measures the other one to be moving at only 0.88c using their own rulers and clocks.

 

[bernhard.rothenstein]

c-u,c+u

It's sometimes called the "closing velocity"--the speed that the distance between two objects is seen to shrink (or grow, if the faster object is moving away from the slower one) as seen in the frame of a third observer. Another example is that if I see two ships moving towards each other, both moving at 0.6c in my frame, then for me the closing velocity is 1.2c (i.e. if they start out 1.2 light years apart as measured by my ruler, it will take them 1 year to meet according to my clocks), even though each ship measures the other one to be moving at only 0.88c using their own rulers and clocks.

Thank you for your answer.Did you take into account that c represnts the speed of light in empty space?
sine ira et studio

 

[JesseM]

Thank you for your answer.Did you take into account that c represnts the speed of light in empty space?
sine ira et studio

Yes, and in every frame the light is measured to move at c. In frame I the light itself obviously moves at c, and although frame I measures the "closing velocity" between the light beam and R to be (c-u), in R's own rest frame the light is approaching him at c, not c-u.

 

[bernhard.rothenstein]

Yes, and in every frame the light is measured to move at c. In frame I the light itself obviously moves at c, and although frame I measures the "closing velocity" between the light beam and R to be (c-u), in R's own rest frame the light is approaching him at c, not c-u.

I started to think about the problem after reading
"Quantization, Doppler shift and invariance of the speed of light:some didactic problems and opportunities" by G.Margaritondo Eur.J.Phys. 16 (1995) 169-171 by G.Margaritondo.
I invite to a discussion about his conclusions.

 

[bernhard.rothenstein]

c-u what is that?

I assume you mean that R coincides with S initially and is moving AWAY from S at speed u. From t= 0 to Tr, S moves a distance uTr so the light beam must travel that distance: c(Tr- Te)= uTr.


Since everything is done in S's frame of reference, yes. Of course, in R's frame of reference the calculation will be quite different.

Speed is defined as traveled distance per travel time. From the equation above we obtain
c-u=cTe/Tr. Is c-u in accordance with the definition above?
Sine ira et studio