# Where is the mistake?

1. Oct 24, 2007

### Andreatn

Can somebody explain the reason why the following reasoning is wrong?

Considering a source of electromagnetic waves at rest with respect to a reference frame, plus an observer who is approaching it with speed v, he would experience a Doppler Effect.

To measure the value of this effect we can use the equation that is in every specific book: if f is the frequency issued from the source, c the speed of light, the frequency F measured by the

observer will be: F = ( 1 – v / c ) f

that is the same as F = [ ( c – v ) / c ] f

Because the observer is moving in the opposite direction to light propagation, we must calculate as

follow: F = { [ c – ( - v ) ] / c } f

and F = [ ( c + v ) / c ]f

if the speed of light is the same for all the observers, than c + v = c

hence: F = ( c / c ) f

and: F = f

thus, the observer who is approaching the source could not verify a Doppler Effect, but the experience shows that this is not true.

2. Oct 24, 2007

### Ich

Your reasoning reminds me of a certain member of some german forums.
1. You got the Doppler effect wrong, you should use the relativistic one. But that doesn't change much.
2. c+v trivially equals c+v, as long as you measure both velocities in the same reference frame. This is the case in the derivation of the doppler effect. A statement like c+v=c is definitely wrong, unless one states explicitly that the "+" sign is meant to perform the "relativistic velocity addition". Even then, it is extremely sloppy and promotes misunderstanding.
The relativistic velocity addition is only appropriate when velocities as measured in different reference frames have to be connected. This operation is quite different from the algebraic addition and should not be confused with it.

3. Oct 24, 2007

### Demystifier

The Einstein ideas were revolutionary, but still not that revolutionary. They changed physics, but still not mathematics. Thus, c+v is not equal to c.

4. Oct 25, 2007

### Andreatn

I can not agree

Dear Ich,

1.Yes, I did not use the relativistic formula for the Doppler effect, but it is negligible.

2. If c+v=c is not true, than we will have an inertial observer who measure a speed of light more than c.
I do not well understand what you exactly mean with "relativistic velocity addition", but
I do not think that Michelson e Morley was thinking at this kind of addition when they discover that c+v=c because they did not yet know the relativity theory and their reasoning was referred to one frame only, like in our case.

5. Oct 25, 2007

### JesseM

The problem here is that your notation is confused, its not clear what those symbols are supposed to represent exactly. If you're trying to relate the measured velocities of multiple different frames, it needs to be expressed more clearly. For example, say w represents the velocity of an object in one frame, and u represents the velocity of the same object in a different frame which is moving at speed v relative to the first one (in the same direction that the object is moving). Then the relativistic velocity addition formula says that u = (w + v)/(1 + wv/c^2), which means that if w=c, then u=c as well. On the other hand, if w and u just represented the velocity of two objects moving in opposite directions in a single frame, then the rate that the distance between them was increasing would just be w + u, and it is quite possible that the distance between two objects moving apart can increase faster than light in a given frame, even though neither object is moving faster than light in that frame. So, it's critical to define exactly what all your terms represent, and which frame they're supposed to be measured in.

6. Oct 25, 2007

### Ich

They discovered that c=c, nothing more. Their apparatus was at rest in their frame, v=0. They had no theoretical framework to explain this result, as they expected to detect the aether. So I'd say it's largely irrelevant what they thought.

To recapitulate what you hear from three members:
1. there is a mathematical operation called relativistic velocity addition which combines velocities that are defined in different reference frames.
2. One should never use the "+"-symbol to refer to this operation, because it naturally leads to that kind of confusion that we see here.

So, whenever you read "c+v" in a sensible text, it means "c+v" and nothing more. As Demystifier said, nobody changed the meaning of the "+"-symbol.

7. Oct 26, 2007

### Andreatn

distance between them was increasing would just be w + u, and it is quite possible that the distance between two objects moving apart can increase faster than light in a given frame, even though neither object is moving faster than light in that frame.
Dear JesseM,
I was a little be confused about the relativistic velocity addition formula because it was exactly what I used to calculate the result, but I avoid to write it completely.
I thought that it was clear that if I write c+v=c I did not use the Galilean formula to add the velocities.
I can not agree with you, because Einstein clearly affirms that his theory is based on the fact that the speed of light is the same for all the inertial observers.
No matter how many referring frames are involved.
When Michelson and Morley discover that c+v=c they had the result with reference to a frame only.
The problem is clear and simple: here we have an inertial observer who must measure a speed of light faster than c to get the Doppler effect.
To be (c+v=c) or not to be?
If you affirm that an inertial observer can measure a speed of light faster than c you say that the theory of relativity is wrong.

8. Oct 26, 2007

### Andreatn

Dear demystifier,
you are right, I use the expression c+v =c to directly indicate the result of the relativistic addition of velocities formula and it is not exactly correct.
I must write : c=(c+v)(1+cv/c^2).

9. Oct 26, 2007

### Andreatn

10. Oct 26, 2007

### JesseM

Can not agree with which statement of mine? I agree that the speed of light is the same for all inertial frames, nothing I said contradicted that.
Actually, they specifically repeated the measurement at different times of year so that the Earth would have different rest frames in each measurement due to its orbit. The wikipedia article gives more detail.
I don't really understand your argument. For example, why do you say that
But then immediately go on to say:
How are these two expressions the same? The brackets don't change anything, the only thing you've done is replaced the 1 with c, why did you do that?

It would also be helpful if you could restate your argument using the correct equation for the relativistic Doppler effect:

$$F = f \sqrt{\frac{1 + v/c}{1 - v/c}}$$

11. Oct 26, 2007

### Ich

HI JesseM,

They are, you overlooked the brackets.

Hi Andreatn,
Nothing, if you accept the principle of relativity.
We don't. Every theory based on the relativity principle (e.g. Ritz) would explain it.
It's just that there are different types of experiments that such a theory must also explain. That leaves us with SR.
How could that be clear? You redefined the meaning of "+" to generate the confusion you eventually got lost in.
No, there is no such observer. There is one who measures a frequency shift according to the formula JesseM gave. This observer calculates, amongst other things, the expression "c+v", but never measures light with speed c+v.

12. Oct 26, 2007

### Andreatn

Sorry boys,
this week end I am busy and I will have not time enough to prepare the anwers,but, be sure, I will send them.
Dear Ich, if you think so you will have a lot of things to explain.
Of course, I agree with the principle of relativity, but I mean that M&M did not think that their apparatus was at rest in its frames.

13. Oct 26, 2007

### Andreatn

O. K. JesseM,
I will explain better the problem and I promise that I will use the mathematics formulas with more attention, but you must promise to me that you will mind more at the fact that is physically behind the formula.
Have a nice week end

14. Oct 26, 2007

### JesseM

Well, one way of considering what is "physically behind the formula" is to try to derive it from relativity:

Consider a source which is sending out pulses every t seconds, in its own rest frame. In my frame, it is approaching with velocity v, and because of time dilation, it's only sending out pulses once every $$\frac{t}{\sqrt{1 - v^2/c^2}}$$ seconds. But also, because it is moving towards me, each successive pulse is emitted at a shorter distance from me than the previous one--if the velocity is v, then in a time of $$\frac{t}{\sqrt{1 - v^2/c^2}}$$ seconds the source must get closer by a distance of $$\frac{vt}{\sqrt{1 - v^2/c^2}}$$. If each pulse is moving towards me at a speed of c, then if each pulse's distance to get to me is decreased by $$\frac{vt}{\sqrt{1 - v^2/c^2}}$$, then that will save each successive pulse a time of (that distance)/c in making the trip (as compared to the time it took the previous pulse to reach me), or $$\frac{vt}{c \sqrt{1 - v^2/c^2}}$$.

So, even though the actual time between pulses being emitted in my frame is $$\frac{t}{\sqrt{1 - v^2/c^2}}$$, the time T between successive pulses reaching my position is instead given by:

$$\frac{t}{\sqrt{1 - v^2/c^2}} - \frac{vt}{c \sqrt{1 - v^2/c^2}}$$

This simplifies to:

$$T = \frac{ct - vt}{c \sqrt{1 - v^2/c^2}}$$

which can be simplified by replacing the c outside the square root in the denominator with a c^2 inside the square root:

$$T = \frac{ct - vt}{\sqrt{c^2 - v^2}}$$

And (c^2 - v^2) = (c+v)(c-v), so this simplifies to:

$$T = \frac{t (c - v)}{\sqrt{c + v} \sqrt{c - v} }$$

which simplifies to

$$T = t \frac{\sqrt{c - v}}{\sqrt{c + v}}$$

So, this is the time between pulses that I will see. Of course, the time between pulses t in the source's rest frame is just 1 over the frequency f in the source's rest frame, so this is equivalent to:

$$T = \frac{1}{f} \frac{\sqrt{c - v}}{\sqrt{c + v}}$$

And the time T between pulses that I see in my frame is just 1 over the frequency F in my frame, so this gives:

$$F = f \frac{\sqrt{c - v}}{\sqrt{c + v}}$$

...which is the relativistic Doppler equation for a source moving towards you. Note that $$\frac{\sqrt{c - v}}{\sqrt{c + v}}$$ can be rewritten as $$\frac{(\sqrt{1 - v/c})/(\sqrt{c})}{(\sqrt{1 + v/c})/(\sqrt{c})}$$, and canceling out the factor of $$1/\sqrt{c}$$ from numerator and denominator gives $$\frac{\sqrt{1 - v/c}}{\sqrt{1 + v/c}}$$ or just $$\sqrt{\frac{1 - v/c}{1 + v/c}}$$, so this is the same as the equation:

$$F = f \sqrt{\frac{1 - v/c}{1 + v/c}}$$

Last edited: Oct 26, 2007
15. Oct 26, 2007

### Ich

You mean that they thought that the apparatus was moving relative to itself??
If you think so you will have a lot of things to explain.

16. Oct 26, 2007

### JesseM

Ah, I misunderstood. I thought when he wrote F = ( 1 – v / c ) f that he meant $$F = \frac{1 - v}{c} f$$ rather than $$F = (1 - \frac{v}{c}) f$$.

17. Oct 27, 2007

### Mentz114

The M&M experiment was set up to test if there exists a luminiferous aether. It was in no way a test of SR because it was done in 1887, a long time before SR. The only thing that experiment can tell us is that there is no aether of that type.

The OP is using the wrong formula - JesseM has derived it properly -

$$F = f \sqrt{\frac{1 - v/c}{1 + v/c}}$$

18. Oct 29, 2007

### Andreatn

wrong idea

But also, because it is moving towards me, each successive pulse is emitted at a shorter distance from me than the previous one--

Dear JesseM,

this is a wrong idea, because a spring of electomagnetic wawes is always in tha middle of the emitted radiation even if it is moving, according with the M&M experiment result
( Bertrand Russel clearly affirm it in his book : ABC of Relativity. I can not give more details about this book, because I only have the Italian version)
Please also see the replay that I will send to Ich.
So the pulses are always emitted at the same distance, even if the source is considered in motion.
It is not a mistake of yours because this wrong idea is largely shared among the scientists, like the figure on the website, that you suggest me to visit to find the relativistic Doppler effect formula, clearly shows.

19. Oct 29, 2007

### pervect

Staff Emeritus
Jesse's end result is a standard (and correct) textbook result.

The thread has moved from a "please point out my error" thread to a claim that standard textbook results are wrong. I don't think the thread is likely to get back on track (to something that's acceptable to our written guidelines), so I'm locking it.