# Calculating different time frames

I have a question
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??

phinds
Gold Member
2021 Award
Have you made any attempt to figure this out yourself?

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

phinds
Gold Member
2021 Award
Have tried time dilation. But i dont think that is the correct way to do it.

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

PeterDonis
Mentor
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.

I have a question
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??
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 Δt0, v is imaginary. Unless I screwed up the algebra.

I have a question
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??
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 Δt0, v is imaginary. Unless I screwed up the algebra.
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?

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

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 Δt0, v is imaginary. Unless I screwed up the algebra.

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

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

phinds
Gold Member
2021 Award
I took this approach. But v comes out to be a little less then c.
Uh ... did you think it should come out at MORE than c?

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.

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.

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

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

PeterDonis
Mentor
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?

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.

jbriggs444
Homework Helper
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.

Ibix
Well i think that the speed should not come anywhere near the speed of light.
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 have no idea as to why this abnormality is produced from time dilation formula.
I'm not seeing anything abnormal. Perhaps you can explain why you don't expect a high velocity?

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

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.

Ibix
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.

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)

jbriggs444
Homework Helper
In fractions , i am getting this ; -
v = [c^2 - (1/(4(c)^2)) ]^(1/2)
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.

Ibix
In fractions , i am getting this ; -

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

jbriggs444
I took this approach. But v comes out to be a little less then c.
v is imaginary if the moving clock moves faster than your clock. Look below. Δt = 1s and Δt0= 100s.

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

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

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

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

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

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

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

Now plug in Δt0 = 100 and Δt = 1.

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

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?

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.

For a time dilation factor of 2 ;
V* = (0.886)c

For time dilation factor of 100;
V = (0.999)c

Now, there is not a very large difference b/w V* and V. But the difference in the Dilation factor is too big. This is the abnormality that i am getting.

v is imaginary if the moving clock moves faster than your clock. Look below. Δt = 1s and Δt0= 100s.

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

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

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

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

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

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

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

Now plug in Δt0 = 100 and Δt = 1.

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

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.

I know that this is a hypothetical situation. But assume this situation in the following way.

Suppose time for the moving person is normal. But time for the stationary person has slowed down. Its the opposite of time dilation, where the time for the moving person has speed up relative to the time of stationary person or vica versa.

I know that this is a hypothetical situation. But assume this situation in the following way.

Suppose time for the moving person is normal. But time for the stationary person has slowed down. Its the opposite of time dilation, where the time for the moving person has speed up relative to the time of stationary person or vica versa.
When you do that the speed is imaginary, as I just showed. When two clocks are in relative motion, whichever clock is looked at from a distance will ALWAYS move slower. Or in the common vernacular, “moving clocks move slowly.” Obviously that phrase misses some subtleties. But, ignoring the grim details, when you are right next to your clock and at rest with respect to it, and you are looking at a far away clock that is moving, you will measure the moving clock to be ticking slower than your clock.

And just FYI: when the other person looks at YOUR clock, s/he sees your clock moving slower. Because s/he is at rest with respect to his/her own clock, and to them, YOU and your clock are the ones moving.

If you want to open the can of worms go ahead, but you can’t do it half way.

To reiterate, here is time dilation solved for v. Plug in values for t and t0 and see what you get. Here t0 is the moving clock.

$$v = c \sqrt{ 1 - \frac{t_0^2}{t^2}}$$

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When you do that the speed is imaginary, as I just showed. When two clocks are in relative motion, whichever clock is looked at from a distance will ALWAYS move slower. Or in the common vernacular, “moving clocks move slowly.” Obviously that phrase misses some subtleties. But, ignoring the grim details, when you are right next to your clock and at rest with respect to it, and you are looking at a far away clock that is moving, you will measure the moving clock to be ticking slower than your clock.

And just FYI: when the other person looks at YOUR clock, s/he sees your clock moving slower. Because s/he is at rest with respect to his/her own clock, and to them, YOU and your clock are the ones moving.

If you want to open the can of worms go ahead, but you can’t do it half way.

Well i think. This is enough discussion for this topic. Clearly, this question is beyond relativity. So if i have to solve it, i would have to make a new branch of physics with completely different ideas from Einstien’s relativity. And i cannot creat this branch.