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If an object is traveling at a certain speed and 1 sec in our time is equivalent to 100 sec in the objects frame of reference. Then how fast is the object traveling??

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- Thread starter Dhruv007
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If an object is traveling at a certain speed and 1 sec in our time is equivalent to 100 sec in the objects frame of reference. Then how fast is the object traveling??

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Have you made any attempt to figure this out yourself?

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Have tried time dilation. But i dont think that is the correct way to do it.

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Show your work.Have tried time dilation. But i dont think that is the correct way to do it.

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Ibix

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You need to be careful about precisely what you mean by the question you are asking, but this method will give you the correct answer under the obvious interpretations (possibly with a couple of caveats). As phinds says - post your working. Note that you can use LaTeX for maths - see link below the comment box if you aren't familiar with it. Why you want to know would also help to see if you are using an appropriate approach.Have tried time dilation. But i dont think that is the correct way to do it.

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Have tried time dilation.

As @phinds has said, please show your work. Without the details of what you have attempted, we don't know where your confusion is or how to help you resolve it.

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Presumably, you are right next to your clock and the object is right next to its clock. So why couldn't you just solve for v in this?

If an object is traveling at a certain speed and 1 sec in our time is equivalent to 100 sec in the objects frame of reference. Then how fast is the object traveling??

$$Δt = \frac{Δt_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Also this seems kind of impossible, because according to your frame of reference THE OBJECT is the one in motion, and hence it should have a slower clock than yours.

EDIT- Yep, if you put 1 in for Δt and 100 in for Δt

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Also if you flip the numbers to better reflect reality (the moving clock will run slower, so you experience 100 seconds while the object experiences 1 second, according to your measurement), you do in fact get a reasonable number for v (it's near light speed, but still slower, and definitely not imaginary ).Presumably, you are right next to your clock and the object is right next to its clock. So why couldn't you just solve for v in this?

If an object is traveling at a certain speed and 1 sec in our time is equivalent to 100 sec in the objects frame of reference. Then how fast is the object traveling??

$$Δt = \frac{Δt_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Also this seems kind of impossible, because according to your frame of reference THE OBJECT is the one in motion, and hence it should have a slower clock than yours.

EDIT- Yep, if you put 1 in for Δt and 100 in for Δt_{0}, v is imaginary. Unless I screwed up the algebra.

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Presumably, you are right next to your clock and the object is right next to its clock. So why couldn't you just solve for v in this?

$$Δt = \frac{Δt_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Also this seems kind of impossible, because according to your frame of reference THE OBJECT is the one in motion, and hence it should have a slower clock than yours.

EDIT- Yep, if you put 1 in for Δt and 100 in for Δt_{0}, v is imaginary. Unless I screwed up the algebra.

I took this approach. But v comes out to be a little less then c.

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Ibix

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That sounds reasonable. What answer did you get exactly? And why are you doing this? What scenario do you have in mind? It's perfectly possible that you are doing maths correctly, but the maths does not relate to the scenario you are actually considering.I took this approach. But v comes out to be a little less then c.

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Uh ... did you think it should come out at MORE than c?I took this approach. But v comes out to be a little less then c.

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Uh ... did you think it should come out at MORE than c?

Well according to me. The answer should not be close to speed of light at all.

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That sounds reasonable. What answer did you get exactly? And why are you doing this? What scenario do you have in mind? It's perfectly possible that you are doing maths correctly, but the maths does not relate to the scenario you are actually considering.

Yep. The math is correct but the answer is not reasonable. I am taking this scenario out of curiosity.

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Nugatory

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If the time dilation factor is 100 then the relative speed must be very close to ##c##. I don't understand why you're thinking it shouldn't be.Well according to me. The answer should not be close to speed of light at all.

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Ibix

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The answer is perfectly reasonable. If the relative speed were low, you'd have nearly Newtonian physics, and hence nearly no time dilation. You need a high speed for relativity to be important.Yep. The math is correct but the answer is not reasonable. I am taking this scenario out of curiosity.

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Well according to me. The answer should not be close to speed of light at all.

What do you think the answer should be, and why?

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What do you think the answer should be, and why?

Well i think that the speed should not come anywhere near the speed of light. But i am getting a speed of 299 792 456.000000 m/s.

I have no idea as to why this abnormality is produced from time dilation formula. Maybe time dilation is not applicable in this scenario. Does anyone have an idea about how to solve this using newtonian mechanics.

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jbriggs444

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Well i think that the speed should not come anywhere near the speed of light

That's what you think. But not why you think it.

- #19

Ibix

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Whyever not? You want a time dilation factor of 100. That means ##v^2## needs to be within one part in ##100^2## of ##c^2##, or about v=0.99995c, which I assume is what your figure is.Well i think that the speed should not come anywhere near the speed of light.

I'm not seeing anything abnormal. Perhaps you can explain why you don't expect a high velocity?I have no idea as to why this abnormality is produced from time dilation formula.

Time is absolute in Newtonian physics, so this would be impossible.Does anyone have an idea about how to solve this using newtonian mechanics.

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Whyever not? You want a time dilation factor of 100. That means ##v^2## needs to be within one part in ##100^2## of ##c^2##, or about v=0.99995c, which I assume is what your figure is.

I'm not seeing anything abnormal. Perhaps you can explain why you don't expect a high velocity?

The abnormality is that “No matter what the time dilation factor is, the value of v always comes so close to c”. I tried the same scenario with a dilation factor of 2 and still got v as 299,792,457.999999993329 m/s.

Now how is this possible. V is almost same for both the scenarios.

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Ibix

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That's wrong. Please show your working.The abnormality is that “No matter what the time dilation factor is, the value of v always comes so close to c”. I tried the same scenario with a dilation factor of 2 and still got v as 299,792,457.999999993329 m/s.

Suggestion - don't bother working out the velocity in terms of m/s, just give it as a fraction of c (e.g. 0.99959c, as I did). It's a lot easier to keep track of the meaning.

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That's wrong. Please show your working.

Suggestion - don't bother working out the velocity in terms of m/s, just give it as a fraction of c (e.g. 0.99959c, as I did). It's a lot easier to keep track of the meaning.

In fractions , i am getting this ; -

v = [c^2 - (1/(4(c)^2)) ]^(1/2)

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jbriggs444

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There is an algebra error there. What you have written is: ##v=\sqrt{c^2-\frac{1}{4c^2}}## That formula is not even dimensionally consistent. Please show your work.In fractions , i am getting this ; -

v = [c^2 - (1/(4(c)^2)) ]^(1/2)

- #24

Ibix

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The units don't make sense on that - you've got a 1/c added to a c, so you've messed up the algebra somewhere. You correctly stated the time dilation formula above:$$\frac{t}{t'}=\frac{1}{\sqrt{1-v^2/c^2}}$$, and you want ##t/t'=2##. Solving that shouldn't be too hard.In fractions , i am getting this ; -

v = [c^2 - (1/(4(c)^2)) ]^(1/2)

- #25

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v is imaginary if the moving clock moves faster than your clock. Look below. Δt = 1s and ΔtI took this approach. But v comes out to be a little less then c.

$$Δt = \frac{Δt_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$

$$Δt^2 = \frac{Δt_0^2}{1 - \frac{v^2}{c^2}}$$

$$1 - \frac{v^2}{c^2} = \frac{Δt_0^2}{Δt^2}$$

$$ - \frac{v^2}{c^2} = \frac{Δt_0^2}{Δt^2} - 1 $$

$$ \frac{v^2}{c^2} = 1 - \frac{Δt_0^2}{Δt^2}$$

$$ v^2 = c^2 (1 - \frac{Δt_0^2}{Δt^2} )$$

$$ v = c \sqrt{ 1 - \frac{Δt_0^2}{Δt^2}}$$

Now plug in Δt

$$ v = c \sqrt{ 1 - \frac{100^2}{1^2}}$$

What is v?

This is an impossible scenario because you have set it up so that the moving clock (the object) is ticking faster than the rest clock (your clock). That is why v is imaginary.

Edit: If instead the MOVING clock is 1 second and YOUR clock is 100 seconds, you get the reasonable answer of just less than the speed of light. It’s the same algebra, but when you get to this step swap your delta t’s.

$$ v = c \sqrt{ 1 - \frac{Δt_0^2}{Δt^2}}$$

$$ v = c \sqrt{ 1 - \frac{1^2}{100^2}}$$

Now what do you get?

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