If a mass is moving with speed of light does it converts into energy of remains as mass it self?As per relativity theory if a mass moves with speeds comparable to speed of light then the time on the mass will be moving slower when compared to the time from where it has been started from.So if you use relativity theory as the mass approaches its speed to speed of light ,then the time or watch on the mass stops or the mass gains infinite time or does it convert to energy or what else can happen?can anyone give a try for this question?
[Physics] what if an object moves with speed of light,does it converts into energy or mass
relativity
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Beginners to special relativity generally learn the equation for time dilation:
$$ t' = \frac{t}{\gamma} \tag{1} $$
where $\gamma$ is the Lorentz factor:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
For any $v > 0$ the Lorentz factor is greater than one and hence $t' < t$, which is what we mean by time dilation.
And this is all very well, but equation (1) is a special case that applies only for uniform unaccelerated motion, and it's absolutely no use for understanding the twin paradox. Indeed, the twin paradox only arises because people mistakenly apply equation (1) when it isn't appropriate. To understand the twin paradox properly we have to go a bit deeper into how special relativity works.
Special relativity is a geometrical theory of spacetime, just like general relativity in fact though considerably simpler. The basic equation for special relativity is the Minkowski metric:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$
Suppose I set up a coordinate system $(t, x, y, z)$ with myself stationary at the origin and the time $t$ measured with my clock and the distances $x$, $y$ and $z$ measured with my rulers. What the Minkowski metric says is that if I observe an object to move a distance $(dx, dy, dz)$ in a time $dt$ then the proper time $d\tau$ can be calculated using equation (2).
Why would we do this? Well firstly the proper time is an invariant i.e. every observer in any coordinate frame moving, rotating, accelerating or whatever will calculate the same value for the proper time $d\tau$. Secondly the proper time is just the time shown on a clock carried by our moving object.
The invariance of the proper time is an assumption and actually it's the key assumption in special relativity. You just have to accept that's how SR works. However we can easily prove the second point i.e. that the proper time is the time shown on the moving object's clock. In the moving object's rest frame it is stationary so $dx = dy = dz = 0$ and therefore equation (2) simplifies to:
$$ c^2d\tau^2 = c^2dt^2 $$
or just:
$$ d\tau = dt $$
So in the object's rest frame the proper time $d\tau$ is equal to the time measured by the object's clock $dt$. But we've just said that $d\tau$ is an invariant and is the same for all observers. That means for all observers the proper time is the time shown on the moving objects clock.
The point of all this is that we now have a way to calculate time dilation because we can use equation (2) to calculate the proper time. Let's do this for the simple case of uniform unaccelerated motion and check we get equation (1). We'll assume the object passes us at time zero with velocity $v$, and for convenience we'll assume it's moving along the $x$ axis so $dy = dz = 0$. The equation (2) becomes:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$
But if the object is moving at velocity $v$ this means that $dx/dt = v$ because that's what we mean by velocity. So $dx = vdt$, and substituting this into the equation about we get:
$$ c^2d\tau^2 = c^2dt^2 - v^2dt^2 $$
and a quick rearrangement gives:
$$ d\tau^2 = dt^2(1 - v^2/c^2) $$
or:
$$ d\tau = \frac{dt}{\frac{1}{\sqrt{1 - v^2/c^2}}} $$
which is just equation (1).
This is a good point to stop and summarise where we've got to. We've stated that the Minkowski metric, equation (2) above, is the core equation in special relativity, and that it can be used to calculate the proper time $d\tau$, which is just the time measured by the moving object's clock. And this is what you asked, i.e. how do we calculate how much time has elapsed for the moving object. So in principle at least we've answered your question.
But can we do the calculation for an accelerating observer like the twin? And the answer is that yes we can. The maths will get harder by the same basic principle applies. We'll do it for a simplified case. Suppose at time zero our object starts accelerating at a constant (in our frame) acceleration $a$. Then in our frame the distance is given by the equation we all learned in school:
$$ x = \tfrac{1}{2}at^2 $$
and differentiating gives:
$$ dx = atdt $$
and as before we substitute this into the Minkowski metric to get:
$$ c^2d\tau^2 = c^2dt^2 - a^2t^2dt^2 $$
or:
$$ d\tau = dt\sqrt{1 - \frac{a^2}{c^2}t^2} $$
Integrating this turns out to be rather sticky, so we'll assume that $at \ll c$ in which case we can approximate the square root as;
$$ d\tau \approx dt\left(1 - \frac{a^2}{2c^2}t^2\right) $$
And we can integrate this to get the elapsed time for the accelerating object:
$$ \tau \approx t\left(1 - \frac{a^2}{6c^2}t^2\right) $$
And we find that $\tau \lt t$ so the time for the accelerating object is dilated - the accelerating twin ages less than us.
There are some misconceptions in your question, and you are really asking two different things.
First, remember that if you are on a spaceship moving at high speeds, Earth people will measure your watch as moving slower than theirs and you will measure their watches as moving slower than yours! Relativity works boths ways, that's why it's called relativity.
A light year is a unit of distance, just like a meter or a mile. It is defined as the distance that light travels in a year; this definition implicitly assumes that the same observer will measure both distance and time, and that this observer is inertial. One important point is that you do not actually have to go out and measure a light year to know what it is! Earth's velocity doesn't matter, because a light year is defined assuming you are in some inertial frame, it doesn't matter which one.
If distance and time are measured by different observers, things will change: If one of the observers gets on a rocket ship and travels at $0.9999c$ for a year proper time, someone else standing on Earth will measure the travelled distance to be much more than $0.9999$ light years (around 70 light years, in fact).
This is also the answer for the last part of your question. If you travel 5 light years (as measured from Earth!) at a speed practically equal to $c$, when you get there your clocks will read much less than 5 years.
Best Answer
One of the things that frequently trouble beginners to special relativity is that all velocities are relative. We can accelerate protons to $0.999999991c$ in the Large Hadron Collider, but from the proton's perspective it is at rest and it's everything outside the collider that's moving. So we can be confident that nothing happens to objects moving at near light speed velocities, because nothing has happened to us even though we are moving at light speed velocities relative to other observers.
It is certainly true that from our perspective time slows down for a fast moving object e.g. an astronaut on a fast moving rocket. However the astronaut would still observe time passing at the normal rate, and indeed they would observe our time to have slowed.
All this seems utterly bizarre, but that's only because we don't usually observe things moving that fast so we aren't used to the effects it has. As you study relativity and get used to it you'll find the effects like time dilation seem more natural (then you start studying general relativity and it all gets weird again :-).