Does c-u mean that c composes with u?

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In summary, the conversation discusses a source of light S and an observer R moving with speed u. The observer R is initially located in front of S and receives a first light signal emitted by S. S then emits a second light signal at a later time Te, which is received by R at a time Tr. The conversation then goes on to discuss the equation c(Tr-Te)=uTr and its meaning in the two frames of reference. The concept of "closing velocity" is also introduced as the speed at which the distance between two objects appears to shrink or grow in the frame of a third observer. Finally, the conversation raises a question about the definition of speed and its relation to the equation c-u=cTe/Tr.
  • #1
bernhard.rothenstein
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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
 
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  • #2
bernhard.rothenstein said:
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.
 
  • #3
c-u

HallsofIvy said:
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?
 
  • #4
bernhard.rothenstein said:
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.
 
  • #5
c-u,c+u

JesseM said:
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
 
  • #6
bernhard.rothenstein said:
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.
 
  • #7
JesseM said:
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.
 
  • #8
c-u what is that?

HallsofIvy said:
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
 

Related to Does c-u mean that c composes with u?

1. What is the meaning of c-u in composition?

The notation c-u refers to the mathematical concept of function composition, where c is composed with u. This means that the output of u becomes the input of c.

2. How is c-u different from c(u)?

C-u and c(u) are two different notations for the same concept of function composition. The c-u notation is more commonly used in mathematics, while c(u) is often used in computer science.

3. Can c and u be any type of functions?

Yes, c and u can be any type of functions, as long as the output of u is a valid input for c.

4. What is the result of c-u if u and c are inverse functions?

If u and c are inverse functions, then the result of c-u would be the identity function, which returns the input value as the output.

5. How is function composition related to function application?

Function composition is a way to combine two functions to create a new function, while function application is the process of using a function to calculate an output value for a given input. Function composition can be seen as a generalization of function application.

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