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Q & A: Does gravity affect radioactive decay rates?

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Most recent answer: 02/10/2015
Q:
Does a strong gravitational field affect radioactive decay rates? For example, would the actual 1/2 life of Plutonium 238 measure the same for a sample the size of a grapefruit as that of a sample the size of the earth? My understanding of Relativity theory would assert that the notional decay rate of 87 years for this element would be slowed down by the more powerful gravitational field of the larger sample. Yet, radioactivity seems to me to be a random process governed by quantum mechanics and not Relativity. Thank you for any insights.
- PHILIP LANKS (age 65)
HEDGESVILLE
A:

Dear Philip,

Your intuition is correct: gravity does not affect radioactive decay rates.

LeeH

 

As seen from a remote location, being "downhill" in a field does slow all rates, including radioactive decay rates. Locally, since all the rates change together, there's no effect. The effect for something on the Earth's surface compared to something far away is only about a part per billion. A lot more dramatic things than that happen for a large chunk of plutonium.  Mike W.


(published on 03/11/2014)

Follow-Up #1: relativity and radioactive decay

Q:
After I submitted my original question I did have misgivings about the example that I used. Maybe any attempt to gather that much Plutonium would result in a fission reaction before it could be done! What I had in mind was that general relativity gravitational time dilation might not apply to radioactive decay rates. I should have asked what happens to a sample of Plutonium 238 when it is put into orbit around the earth. Would the sample decay at the same rate as an equal amount that is left back on earth, even if by a tiny amount? It's sort of like the twins experiment. Thank you for your website. I really find the questions and answers fascinating.
- PHILIP LANKS (age 65)
HEDGESVILLE
A:

Nice  question. In low-earth orbit, the radioactivity would be very slightly slowed down, just like any other twin-clock. In high orbit the General relativistic effects are bigger, and the radioactive decay would be just slightly faster than on the Earth's surface. If any clock behaved differently than all the other clocks, then you could use it to tell whether you were moving, etc. and the principle of relativity would be violated.

Mike W.

 

Remember, the answer depends on the relative velocities of the object and the observer.  If the object is traveling in orbit and you are on the Earth you will observe a small time difference.   If you are traveling in the same orbit as the source then you will not observe any difference in the decay rate.  LeeH


(published on 03/12/2014)

Follow-Up #2: clocks in orbit

Q:
I was trying to grasp how how radioactive decay of two twin samples of a radioactive material (lets say Carbon-14) could be any different if one was measured from Earth and one was in space. If the two samples were traveling in the same perpendicular orbit at the same speed, it seems they should decay at exactly the same perceived rate.It seems the measurable difference would come if either the one in orbit, or the one on earth did not remain perpendicular to the other, or if the one in outer orbit had an elliptical rotation, if in either case it caused a difference in speed measured between the two sources to create a full revolution.Would the radioactive source that took longer to make one complete revolution have decayed at a slower rate or faster rate, relative to the other twin sample??Also, an arithmetic question, how much faster or slower, and how many more/less revolutions would the Carbon-14 in orbit have to make to perceive zero decay relative to the twin sample on Earth?Maybe if we find the answer to that, we unlock the secret to stop the atomic clock from ticking.
- Chris
A:

First, ditch the "it seem"s. Our untrained intuitions are not well evolved to grasp relativistic effects.

For these orbiting clocks there are two effects:

1. Special relativistic, due to relative velocities.

2. General relativistic, due to accelerations and gravity.

In one way, this all becomes simple because everybody is in repeated orbit. You can make repeated comparisons of clocks without any ambiguity from changing signal transmission times. You can thus compare how much time has elapsed on two clocks directly. So everyone here will agree on who's faster and who's slower and by what factor, regardless of which clock they prefer to use.

Let's pick the Earth frame, for convenience. The clock in circular orbit at H above the earth's surface will travel at speed v=sqrt(GM/(R+H)) where G is the universal gravitational constant, M is the mass of the earth, and R is the radius of the earth. That means we see its clock run slow by a factor of sqrt(1-GM/c2(R+H)). It's also up in a gravitational field, which makes us see it run fast by a factor (1+GM((1/R)-(1/R+H)/c2). All calculations are to lowest order, an excellent approximation.

These effects cancel at H=2R. below that the orbiting clock runs slow, above that it runs fast, according to us.

To get a clock to run at zero rate, according to us, you can't get away with any of the small-effect approximations I've used here. Further just sending it into higher orbit doesn't work, that speeds it up slightly. You need somewhere else for it to go very far downhill in a gravitational field. The event horizon of a black hole is just the place. But that's another story.

Mike W.


(published on 02/10/2015)

Follow-up on this answer.